# RLC circuit question

#### Wendy

Joined Mar 24, 2008
21,914
Two polarized caps, back to back facing opposite directions, make a non-polarized cap. The total capacitance is the same as the two caps calculate in series. Just one of the idosyncracies of electronics.

I *think* electrolytic caps (polarized and nonpolarized) have a higher internal resistance, but I'm not sure of that statement.

Joined May 28, 2009
508
Typically trimmer (non-polarized) caps are used in tank circuits so the frequency can be fine-tuned. It is unusual to find an electrolytic in a tank circuit because the charge/discharge rate is much slower than, say a 22pF capacitor...and you will have a hard time finding any polarized caps in the picofarad range. However, I've seen polarized caps in a multi-thousand watt transmitter. They were about the size of a large watermelon, as was the coil.

#### count_volta

Joined Feb 4, 2009
435
Two polarized caps, back to back facing opposite directions, make a non-polarized cap. The total capacitance is the same as the two caps calculate in series. Just one of the idosyncracies of electronics.

I *think* electrolytic caps (polarized and nonpolarized) have a higher internal resistance, but I'm not sure of that statement. Like this? The two negatives somehow cancel each other out or what?

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#### count_volta

Joined Feb 4, 2009
435
I just had a massive revelation. Finally starting to understand what you guys were trying to tell me. I read my circuits book from university, the section on differential equation analysis of RLC circuits.

It seems that parallel RLC and series RLC are complete opposite. In parallel RLC the bigger your resistance is, the LONGER your oscillations will last.

So it says the following. α is the damping coefficient. When you solve the differential characteristic equation you end up with the poles

s1= -α+ sqrt(α^2-ω^2) and s2 = -α- sqrt(α^2-ω^2)

If ω> α the poles are complex and real conjugates and you have an underdamped case (which is what I want) and the response is e^-αt(Acosωt+Bsinωt) in other words it oscillates and decays.

But here is the important part

α= 1/2RC

In other words as R increases, the denominator increases, and hence α decreases

Since its the α which controls how fast the oscillations decay, it means that the bigger the resistance I use in the parallel RLC the longer the oscillations will last.

This translates to, having nothing in the resistor branch i.e. open circuit, will make the oscillations last the max amount of time. In other words don't use a resistor, just have the inductor and capacitor in parallel. And in parallel the current will split somewhat depending on the real life resistance of the capacitor and inductor.

Now I understand what Bill_Marsden was trying to tell me. Last edited:

#### Wendy

Joined Mar 24, 2008
21,914
You are very close. A series resonant LCR circuit doesn't exacty oscillate like the parallel one does. I suppose you could make it happen with some kind of reflection (something out of impedance matching and RF thoery), but it isn't as automatic as it is with the parallel circuit. With the parallel version the energy bounces back and forth, and the conversion between electric field storage and magnetic field storage is pretty much automatic. The current within the LC circuit is quite high.

The series circuit it is the voltage that is quite high. Have you calculated the voltage drop across the cap and the coil in a series circuit being feed its resonant frequency yet? #### count_volta

Joined Feb 4, 2009
435
But according to α= 1/2RC the value of the capacitor also plays a role in how fast the oscillations die out. Bigger values of the capacitor cause α to be smaller and hence the e^-αt to die out slower.

But I guess it doesn't matter. If I use no resistor in parallel it would be α= 1/∞ since R would be an open circuit aka infinite resistance, and the damping coefficient would be 0.

Of course that is 0 for an ideal capacitor and inductor which have no resistance. The inductor especially, is just wire, and long wire at that, so it has some very real resistance.

I wonder if my ohm meter will be able to measure the resistance of the inductor and capacitor? Its important to estimate the current I will pass through them them also, if I use no resistor.

My 100μH inductor has a rating of 2 amp max. If I pass anything larger through it, I might damage it.

So here is an idea for a circuit I will build. It does not have an amp yet, first I want to see what happens without one. The battery will be 6 volts lets say. I will first try to find the complete response of the current through the inductor using Laplace transform, to solve the differential equations easier. That should in fact give me a function which describes exactly what happens to the signal at every instant of its existence.

As you see I am using a 100 ohm resistor to limit the current through the inductor and capacitor during the charging cycle. After I open the switch, the high current will last a very short time and then decay, but with the switch closed during the charging cycle, it will last long enough to possibly damage the inductor which can't handle anything above 2A.

I will report back after I calculate the response of it. Then try to simulate it with Pspice.

Bill_Marsden please look at this post and help me out. Thanks. #### Wendy

Joined Mar 24, 2008
21,914
The drawing is correct.

To be honest, it doesn't make sense to me. The only thing I figure is the "strong" capacitor (forward biased) takes the strain. A back biased capacitor by itself will blow out, making a nice stink bomb, but add the second cap and it can relax a little without blowing out.

It is an old trick in electronics that works. Understanding why something works is important, but not always necessary.

Concerning your drawing, charge the capacitor by itself with the switch, they only current drawn will be to charge the capacitor. If the coil is "empty" it will take the full charge from cap, then return it. You also get to eliminate the resistor. If both are charged at the same time then they won't oscillate very well.

Do you have access to an oscilloscope? You'll be able to see the ringing with it.

#### count_volta

Joined Feb 4, 2009
435
Hey, well I solved for the current in the inductor using laplace to solve the differential equations.

I am very sure that its correct because I rechecked every step. The best proof I have that its correct is the fact that using the formula for the resonant frequency

f= 1/ (2∏sqrt(LC)) gives me 711.7 Hz as my resonant frequency. And also the solution to my differential equation gave me ω= 4472.1 rad/sec when I found the poles, which is 711.7 Hz. So here is the solution. Current in the inductor above is

I(t) = .06cos(4472.1t) amps.

for t>0 when t=0 represents when the switch has been opened.

No big surprise here. The current in the inductor with the switch closed is 60mA. So when the switch opens, the inductor wants to keep the 60mA flowing.

Obviously this result is ideal and undamped. Its not a decaying cosine because I have no resistor in the LC loop. But real inductors and capacitors do have resistance and so does the wire.

So to figure out a real function for the current, I would need to add the resistance of the inductor and capacitor and probably even the wire. So I need to measure them.

Yea I definitely need an oscilloscope. The problem is they are too expensive. I think its worth it though if I want to take my major seriously.

I will report back after I include the resistance of the inductor and capacitor into the circuit and resolve. I should end up with a decaying cosine.

#### count_volta

Joined Feb 4, 2009
435
Concerning your drawing, charge the capacitor by itself with the switch, they only current drawn will be to charge the capacitor. If the coil is "empty" it will take the full charge from cap, then return it. You also get to eliminate the resistor. If both are charged at the same time then they won't oscillate very well.
By the way, when the switch is closed the inductor is being charged only right? I hope I am correct. As we were taught, an inductor at steady state is equivalent to a short circuit, and a capacitor is equivalent to an open circuit. Here the capacitor is in parallel with a short circuit.

My simulator confirms this. When the switch is closed, the voltage in the capacitor is in nano volts and decreasing, and the current in the inductor is 60mA. When the switch is opened, the inductor charges the capacitor.

Since I want to take the current as my output signal that is what I am interested in. At least before I add the amp its the current, maybe later it will be the voltage, whatever is bigger. I should also solve for the voltage of the capacitor for t>0. First we need to measure the resistance of the cap and inductor though.

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#### Wendy

Joined Mar 24, 2008
21,914
By the way, when the switch is closed the inductor is being charged only right? I hope I am correct. As we were taught, an inductor at steady state is equivalent to a short circuit, and a capacitor is equivalent to an open circuit. Here the capacitor is in parallel with a short circuit.
You're right, took me a second to think it through.

#### count_volta

Joined Feb 4, 2009
435
I tried to figure out the resistance of the inductor. The ohmmeter of my multimeter shows 0Ω. It can't pick it up. Perhaps I need something more sensitive. After all the inductor really has resistance right? 2Ω-5Ω or something.

I tried to figure out the resistance by sticking the inductor in series with a resistor to make a voltage divider, after all if the inductor has real resistance there should be some voltage division, tiny but it should still be noticeable.

But nope the resistor has the entire voltage of the source across it. So the inductor is just being treated as wire by the multimeter.

So then I gave that up and just built the circuit on a breadboard. This circuit. So I attached the multimeter to the capacitor in parallel to measure its voltage. I expected to see some AC voltage even if its millivolts or lower. But nope, although the meter did keep alternating between 0.00 and -0.00.

So then I thought the resistor is keeping the inductor from being charged fully when the switch is closed, so I took the resistor out of the circuit and ended up with the same circuit as above but without the resistor.

Okay then I did see something strange. When I closed the switch I got an AC voltage on the capacitor!!!! I put the meter on AC voltage and got 0.160 volts AC and then it decreases slowly.

But when I open the switch the voltage on the capacitor goes to 0 almost instantly. Close the switch and I get the same 0.160 volts AC and it decreases slowly.

What the heck is going on here? LOL.

Bottom line is, analysis of this circuit is impossible without an oscilloscope!!! I will ask the guys in my department lab to let me use one.

Please read what I wrote carefully before commenting. I wrote it out in great detail in order so that someone can help explain what is going on.

In other news I bought a 741 op amp chip which will eventually be the amp for my circuit.

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#### count_volta

Joined Feb 4, 2009
435
Bill_Marsden since you seem interested in my little project, please let me know what you think.

So today I took my circuit to the lab in my engineering department. The same circuit as in the last post. I got it verified, I did build it correctly, so there is no chance that I made any mistake.

We connected my circuit to an oscilloscope. When I open the switch, yes it oscillated!!!! But it lasted mere milliseconds. The plot looked like a decaying sinusoid as it should. The frequency was slightly higher than the calculated resonant frequency of 711 Hz. The scope showed around 900 Hz. Odd. I talked to my professor who specializes in this field, and he said that he is surprised the signal lasted such a short time. He thinks it should have lasted at least a second.

Anyway my question to you
Bill_Marsden is, what would the op amp in your original circuit do to make the signal last longer? I am not exactly sure what the reason for having it there is? #### Wendy

Joined Mar 24, 2008
21,914
What are your power supplies to the op amp? The 741 is pretty old in the tooth, and is finicky about some things. It also doesn't have the best imput impedance, it is a 40 year old design after all. There are better, but I'll have to get back with you on that.

Try adding a variable resistor between the output and the positive input. This is called positive feedback, and is the basis for most oscillators one way or another.

Adjust the gain using the pot, you should be able to extend the ring time, when gain is too high it will oscillate, come out with a pure tone. There will be some clipping on the sine wave on the output of the op amp, but it should look pretty good across the LC circuit.

Capacitors and inductors are pretty low tolerance compared to resistors, it has to do with measuring instruments being not quite so exact, and the fact that the exact value of a component can actually change a little depending on the frequency.

What you are seeing with your experiments (which I highly approve of) is the difference between theory and practical. If you spend long enough on this project you will probably be able to identify all the factors that make it not quite to prediction, this part of electronics is a pretty rigid science, but the number of variables can be large.

#### count_volta

Joined Feb 4, 2009
435
Thanks, but how exactly will the op amp extend the duration of the oscillations? We learned about op amps last semester. You can do many things with them using some formulas and properties of op amps.

I know that the output of the op amp will be the gain multiplied by the input, I can see how this will make the signal larger in magnitude but how will it extend the duration of the signal?

I told my professor that I want to use an op amp to extend the time that the signal oscillates but he could not quite see how that will be helpful. Maybe you can explain?

Thanks

#### Wendy

Joined Mar 24, 2008
21,914
Recieved as a PM

count_volta said:
Hey, could you please explain to me how an op amp will extend the duration of the oscillations? From what I know it will just amplify the magnitude of the signal.

OK, the circuit as is extends the ring duration by the simple expedient of loading the tank circuit as lightly as possible. Any loading on the tank circuit absorbs the energy in the tank circuit, increasing the damping.

Adding positive feedback takes it one step further. A child on a swing is a simple pendulum, another resonant configuration that is similar in concept to a tank circuit. When the child injects energy (by swinging their legs in time with the swing) the oscillations actually increase in value, or remain the same.

The op amp with positive feedback is much the same. It is adding energy to the tank circuit in phase with the natural oscillation.

Something I'll mention in passing. Resistance is not the only loss mechanism (but it is the primary). These configurations are also common with RF frequencies, and losses in there are in the form of the RF escaping the tank.

#### count_volta

Joined Feb 4, 2009
435
Cool thanks, I will read the section on op amps of the All About Circuits book. This is not such an easy thing to understand. Fascinating though. I am learning so much from this simple project. #### Wendy

Joined Mar 24, 2008
21,914
When I was a teen I went through what you're doing now, only with RF. We lived far out in the country, and TV signals were weak (and my Dad loved TV). Whenever I trashed his reception with my latest RF oscillator (made from home made inductors using toliet tube rolls and ocasionally home made capacitors) his standard phrase was "Turn it off". He didn't have to see what I was doing to know I was responsible for the picture going down the toilet.

I built a lot of projects then, but didn't have much money. Car parts were my friend, along with the old color TV that was was scrapped. My brother was a lot like me, we shared that TV. Many a project recycled resistors and capacitors from it.

If you can find a local electronics store (besides Radio Shack) I would recommend it. Mail Order is OK, but it is also a hassle.

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#### count_volta

Joined Feb 4, 2009
435
I looked in my electronics textbook and found the Colpitts Oscillator and Hartley Oscillator. They are made to do exactly what I am trying to do, i.e. keep the oscillations from decaying. It uses capacitors and an inductor. So its basically my original circuit but with a transistor.

I think I will build a colpitts oscillator. It is much easier to understand than the op amp version. A single transistor is easier to understand and analyze. I don't quite understand how the transistor uses positive feedback to keep the oscillations going just yet, but I will definitely figure it out. I do know how a BJT works so it should not be too hard.

Perhaps someone can explain exactly what the transistor does in the Colpitts circuit? I looked it up online but they love to use fancy words without really explaining anything. Same goes for my textbook.

#### Wendy

Joined Mar 24, 2008
21,914
Show the schematic you're using. And turn it off. #### count_volta

Joined Feb 4, 2009
435
Show the schematic you're using. And turn it off. I have not built anything yet. Nothing to turn off. I am just trying to figure out what the transistor does exactly to keep the oscillations going. Here is a good colpitts circuit. I wont necessarily use those same values though. Could you please explain what the transistor is doing, and why are there 2 capacitors?

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